Question: Find $\lim_{x\to\scriptsize\dfrac{\pi}{2}}\dfrac{\sin(2x)}{\cos(x)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{2}$ (Choice B) B $1$ (Choice C) C $2$ (Choice D) D The limit doesn't exist
Substituting $x=\dfrac{\pi}{2}$ into $\dfrac{\sin(2x)}{\cos(x)}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since our expression includes trigonometric functions, let's try to re-write it using factorization and trigonometric identities. Since we have $\sin(2x)$ in our expression, let's rewrite it using its double angle identity $\sin(2x)=2\sin(x)\cos(x)$. $\begin{aligned} &\phantom{=}\dfrac{\sin(2x)}{\cos(x)} \\\\ &=\dfrac{2\sin(x)\cos(x)}{\cos(x)} \gray{\sin(2x)\text{ identity}} \\\\ &=\dfrac{2\sin(x)\cancel{(\cos(x))}}{\cancel{\cos(x)}} \gray{\text{Cancel common factors}} \\\\ &={2\sin(x)}\text{, for }x\neq \{...,- \dfrac{7\pi}{2}, - \dfrac{3\pi}{2}, \dfrac{\pi}{2}, \dfrac{5\pi}{2}, \dfrac{9\pi}{2},...\} \end{aligned}$ This means that the two expressions have the same value for all $x$ -values (in their domains) except for $(2k+1)\dfrac{\pi}{2}$ for any integer $k$, and specifically $\dfrac{\pi}{2}$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{\sin(2x)}{\cos(x)}={2\sin(x)}$ for all $x$ -values in the interval $(-\pi,\pi)$ except for $x=\dfrac{\pi}{2}$. Therefore, $\lim_{x\to \scriptsize\dfrac{\pi}{2}}\dfrac{\sin(2x)}{\cos(x)}=\lim_{x\to \scriptsize\dfrac{\pi}{2}}{2\sin(x)}=2$ (The last limit was found using direct substitution.) In conclusion, $\lim_{x\to\scriptsize\dfrac{\pi}{2}}\dfrac{\sin(2x)}{\cos(x)}=2$.